5. (a) Show that the rotational kinetic energy of a rigid body is given as T=(1)/(2)Iomega ^2 where the terms have their usual meanings. (5 marks)

To show that the rotational kinetic energy of a rigid body is given by \( T = \frac{1}{2} I \omega^2 \), we start by considering a rigid body rotating about a fixed axis. The body can be thought of as being composed of many small particles, each with mass \( m_i \) and located at a distance \( r_i \) from the axis of rotation.<br /><br />The velocity \( v_i \) of each particle due to the rotation is given by:<br /><br />\[ v_i = r_i \omega \]<br /><br />where \( \omega \) is the angular velocity of the body.<br /><br />The kinetic energy \( T_i \) of each particle is:<br /><br />\[ T_i = \frac{1}{2} m_i v_i^2 \]<br /><br />Substituting the expression for \( v_i \):<br /><br />\[ T_i = \frac{1}{2} m_i (r_i \omega)^2 = \frac{1}{2} m_i r_i^2 \omega^2 \]<br /><br />The total kinetic energy \( T \) of the rigid body is the sum of the kinetic energies of all its particles:<br /><br />\[ T = \sum \frac{1}{2} m_i r_i^2 \omega^2 \]<br /><br />Since \( \omega \) is the same for all particles in a rigid body, it can be factored out of the summation:<br /><br />\[ T = \frac{1}{2} \omega^2 \sum m_i r_i^2 \]<br /><br />The term \( \sum m_i r_i^2 \) is defined as the moment of inertia \( I \) of the rigid body about the axis of rotation:<br /><br />\[ I = \sum m_i r_i^2 \]<br /><br />Thus, the rotational kinetic energy becomes:<br /><br />\[ T = \frac{1}{2} I \omega^2 \]<br /><br />This completes the derivation, showing that the rotational kinetic energy of a rigid body is given by \( T = \frac{1}{2} I \omega^2 \).